What is the integral of 1/((sin(x)-cos(x)-1)) ?

The integral of 1/((sin(x)-cos(x)-1)) is ln|tan(x/2)-1|+C

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integral of 1/((sin(x)-cos(x)-1))

\int \frac{1}{( sin ( x ) - cos ( x ) - 1 )} d x

ln tan (\frac{x}{2}) - 1 + C

Solution steps

\int \frac{1}{sin ( x ) - cos ( x ) - 1} d x

Apply u-substitution

: \int \frac{1}{u - 1} d u

= \int \frac{1}{u - 1} d u

Apply u-substitution

: \int \frac{1}{v} d v

= \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= ln ∣ v ∣

Substitute back

= ln tan (\frac{x}{2}) - 1

Add a constant to the solution

= ln tan (\frac{x}{2}) - 1 + C

f (x) = ln (3) x^{2}

\frac{x ^{9} - 6 x x}{x}

\frac{d}{d x} (5^{x, x})

x^{5}

\int sin (t) 1 + cos (t) d t

What is the integral of 1/((sin(x)-cos(x)-1)) ?
The integral of 1/((sin(x)-cos(x)-1)) is ln|tan(x/2)-1|+C